Question

The asymptotic Bode plot of the transfer function
$K/[1+(s/a\left)\right]$ is given in figure. The error in phase angle and dB gain at a frequency of $\omega =0.5$ are respectively.

• A. ${4.9}^{\circ },0.97dB$
• B. ${5.7}^{\circ },3dB$
• C. ${4.9}^{\circ },3dB$
• D. ${5.7}^{\circ },0.97dB$

Answer 1

The correct option is A ${4.9}^{\circ },0.97dB$

Transfer function
$G\left(s\right)=\frac{k}{1+\frac{s}{a}}$
$Puts=j\omega$
$G\left(j\omega \right)=\frac{k}{1+j\frac{\omega }{a}}$
Magnitude of
$G\left(j\omega \right)=|G\left(j\omega \right)|=\frac{k}{\sqrt{1+\frac{{\omega }^2}{a^2}}}$

$PhaseofG\left(j\omega \right)=\angle G\left(j\omega \right)=-ta{n}^{-1}\left(\frac{\omega }{a}\right)$

$at\omega =0.5a$

$|G\left(j\omega \right){|}_{\omega =0.5a}=\frac{k}{1+\sqrt{\frac{(0.5a{)}^2}{a^2}}}=\frac{k}{\sqrt{1.25}}$

in dB
$=20\log k-20log(1.25{)}^{1/2}$
$=20logk-0.97dB$

From the plot, magnitude=20 log k

$\therefore$ error in dB gain=0.97 dB

$\angle G\left(j\omega \right){|}_{\omega =0.5a}=-ta{n}^{-1}\left(\frac{0.5a}{a}\right)={26.56}^{\circ }$

$Sinceitis-{45}^{\circ }/dec\left(negativeslope\right)at\omega =0.5a$

$phase=-{22.5}^{\circ }$

Error in phase of $\angle G\left(j\omega \right){|}_{\omega =0.5a}$

$=-22.5-(-26.56)$

$=+{4.06}^{\circ }\simeq +{4.9}^{\circ }$